Electrical Power Calculator from Voltage and Current

What does this quantity describe?

The electrical power P of a load is the product of the voltage U across it and the current I through it: P = U·I. It indicates how much electrical energy is converted per unit time.

Comparable to a water pipe, where transported power depends on both pressure (voltage) and flow rate (current): doubling either pressure or flow doubles the transferred power, as long as the other quantity stays constant.

Formula and variables

P = U · I

• P = U · I
• U = P / I
• I = P / U

Choose the inputs correctly

Enter two of the three quantities. Power requires voltage and current, voltage requires power and current, current requires power and voltage. For AC loads with a phase shift, P = U·I gives only apparent power; real power additionally requires the power factor cos(φ).

How to use the calculator

Select the target quantity and enter the two known values with units. For AC loads, check whether it is a purely resistive load (e.g. a heating element) or a load with phase shift (e.g. motor, transformer), since the second case requires the power factor to be considered separately.

Worked example

Given U = 230 V and I = 10 A, substitution gives P = 230 V · 10 A = 2,300 W = 2.3 kW, roughly the power of a household fan heater.

Understand the result and units

A power of 2.3 kW means the load converts 2,300 joules of electrical energy each second. Over one hour of operation, that is 2.3 kWh of energy use, relevant for estimating electricity costs.

The SI unit is the watt (W = V·A). Larger powers are given in kW or MW, with 1 kW = 1,000 W. Datasheets for AC equipment often also state VA (volt-amperes) for apparent power, which must be distinguished from real power in watts.

What is the difference between real and apparent power in AC?

For purely resistive loads, voltage and current are in phase, so P = U·I gives real power in watts directly. For motors, transformers and other inductive or capacitive loads, voltage and current are out of phase. The product U·I then only gives apparent power in volt-amperes (VA); the actually usable real power in watts results from multiplying by the power factor cos(φ), which can range between 0 and 1.

Typical applications

The formula is used to size fuses, cables and switches, estimate household appliance energy use, and roughly size heating elements and other resistive loads.

Assumptions, limits and common mistakes

P = U·I holds exactly for DC and for AC with a purely resistive, in-phase load (power factor 1). For inductive or capacitive loads such as motors, transformers, or fluorescent lamps with ballasts, voltage and current are out of phase; real power is then smaller than the simple product U·I, and P = U·I·cos(φ) applies.

Common mistake: For AC loads with phase shift, do not use P = U·I unchecked as real power — without the power factor cos(φ), only apparent power in VA is calculated, which can exceed the actual real power in watts. Also, do not confuse power (watts, an instantaneous quantity) with energy (watt-hours, a time integral).